Horizontal Circular Motion (AQA A-Level Further Maths): Revision Notes

Worked Example: Finding Maximum Speed on a Horizontal Bend

Question: Calculate the maximum speed (in km h⁻¹) for a car of mass 900 kg to round a bend of radius 60 m on a level road.

Part (a): Using maximum frictional force

Given: The maximum frictional force between the tyres and road is 1440 N.

Solution:

Let $v$ be the maximum speed of the car.

At maximum speed, we need to use the horizontal equation of motion. The friction force provides the centripetal force needed for circular motion:

$F = m \times \frac{v^2}{r}$

Substituting the known values: $1440 = 900 \times \frac{v^2}{60}$

Rearranging: $v^2 = \frac{1440 \times 60}{900} = 96$

Taking the square root: $v = \sqrt{96} \text{ m s}^{-1} = \frac{\sqrt{96 \times 60 \times 60}}{1000} = \mathbf{35.3 \text{ km h}^{-1}}$

The normal reaction force $R$ from the ground acts vertically and therefore does not contribute to the centripetal force.

Part (b): Using coefficient of friction

Given: The coefficient of friction between the car's tyres and the road is $\mu = 0.5$. Take $g = 9.8$ m s⁻².

Solution:

Let the maximum speed of the car be $v$.

Step 1: Resolve vertically

The vertical forces must balance: $R = mg \quad \text{...(1)}$

Step 2: Write horizontal equation of motion

The horizontal force provides the centripetal acceleration: $F = m \times \frac{v^2}{60} \quad \text{...(2)}$

Step 3: Apply friction inequality

The frictional force cannot exceed $\mu R$: $F \leq \mu R$

Substituting from equations (1) and (2): $m\frac{v^2}{60} \leq \mu mg$

Simplifying (mass cancels): $\frac{v^2}{60} \leq 0.5 \times 9.8$

$v^2 \leq 30g$

Therefore, the maximum speed without skidding is: $v = \sqrt{30g} = \sqrt{30 \times 9.8} = \mathbf{17.1 \text{ m s}^{-1} = 61.7 \text{ km h}^{-1}}$

Worked Example: Combining Circular Motion with Hooke's Law

Question: A particle of mass $m$ is tied to a fixed point $O$ on a smooth horizontal table by an elastic string. The string has natural length 0.3 m and modulus of elasticity $\lambda = mg$. The particle moves at a constant angular speed of 20 rpm about point $O$. Calculate the extension of the string. Take $g = 9.8$ m s⁻².

Solution:

Let the tension in the string be $T$ N and let the extension be $x$ m.

Step 1: Write equation of motion towards centre

The angular velocity is: $\omega = 20 \text{ rpm} = \frac{20 \times 2\pi}{60} \text{ rad s}^{-1}$

The radius of circular motion is $(0.3 + x)$ m (natural length plus extension).

Using the centripetal force equation: $T = mr\omega^2 = m(0.3 + x)\left(\frac{20 \times 2\pi}{60}\right)^2$

$T = \frac{4\pi^2 m}{9}(0.3 + x) \quad \text{...(1)}$

Step 2: Apply Hooke's Law

For an elastic string: $T = \frac{\lambda}{\text{natural length}} \times \text{extension}$

Substituting $\lambda = mg$: $T = \frac{mg}{0.3} \times x = \frac{mgx}{0.3} \quad \text{...(2)}$

Step 3: Solve simultaneously

Equating equations (1) and (2): $\frac{4\pi^2 m}{9}(0.3 + x) = \frac{mgx}{0.3}$

Simplifying (mass cancels): $\frac{4\pi^2}{9}(0.3 + x) = \frac{gx}{0.3}$

$4\pi^2(0.3 + x) = 30 \times 9.8 \times x$

Expanding and rearranging: $4\pi^2(0.3 + x) = 294x$

The extension is: $x = \mathbf{0.047 \text{ m} = 4.7 \text{ cm (to 2 s.f.)}}$

The vertical forces (reaction $R$ upward and weight $mg$ downward) balance each other and do not contribute to the horizontal centripetal force. Only the tension $T$ in the string provides the centripetal force.